Proceedings of the 44th Symposium on Ring Theory and ...

(3) P is a maximal σ-coessential extension <strong>of</strong> M(i.e. P f ↠ M is σ-coessential extension
<strong>of</strong> M and if <strong>the</strong>re exists an epimorphism I h ↠ P and I h ↠ P ↠ M is σ-coessential <strong>of</strong> M,
<strong>the</strong>n h is an isomorphism.).
(4) P is isomorphic to P σ (M).
Pro<strong>of</strong>. (1)→(2): Let P be σ-projective and P ↠ f M be a σ-coessential extension <strong>of</strong> M.
Consider <strong>the</strong> following diagram.
0→ ker h → P → h I → 0
↘ f ↓ g
M,
where I is σ-projective, g and h are epimorphisms such that gh = f.
Since F σ ∋ f −1 (0) = h −1 (g −1 (0)) ⊇ h −1 (0), it follows that F σ ∋ h −1 (0) = ker h. As f is
a minimal epimorphism and g is an epimorphism, h is also a minimal epimorphism. Since
I is σ-projective, <strong>the</strong>re exists a submodule L <strong>of</strong> P such that P = ker h ⊕ L and L ∼ = I.
As ker h is small in P , P = L, and so P ∼ = I.
(2)→(1): Let σ be an epi-preserving idempotent radical and P be a minimal σ-
projective extension <strong>of</strong> M. Consider <strong>the</strong> following commutative diagram.
P σ (P ) → j P → 0
g ↓ ↓ f
P σ (M) → h M → 0,
where h and j are epimorphisms associated with <strong>the</strong> projective covers <strong>of</strong> M and P
respectively and g is an induced epimorphism by <strong>the</strong> σ-projectivity <strong>of</strong> P σ (P ). Since P is
σ-projective, j is an isomorphism by Lemma 4. As P σ (P ) and P σ (M) are σ-projective,
g is an isomorphism by <strong>the</strong> assumption. By Lemma 1, it follows that P → f M → 0 is a
σ-coessential extension <strong>of</strong> M.
(1)→(3): Let I → g P be an epimorphism. Let P ↠ f M and I ↠ h M be σ-coessential
extensions <strong>of</strong> M such that fg = h. Consider <strong>the</strong> following exact diagram.
I
g ↙ ↓ h
P → M → 0
f
Since f is a minimal epimorphism, g is an epimorphim. As h and f are minimal
epimorphisms, g is a minimal epimorphim. Since F σ ∋ h −1 (0) = g −1 (f −1 (0)) ⊇ g −1 (0), it
follows that F σ ∋ g −1 (0). Since P is σ-projective, 0 → ker g → I → g P → 0 splits, and so
<strong>the</strong>re exists a submodule H <strong>of</strong> I such that H ∼ = P and I = ker g ⊕ H. As ker g is small
in I, I = H ∼ = P , as desired.
(3)→(1): We show that P is σ-projective. Since P ↠ f M is a σ-coessential extension
<strong>of</strong> M by <strong>the</strong> assumption, an induced morphism P σ (P ) → P σ (M) is an isomorphism by
Lemma 1. Consider <strong>the</strong> following commutative diagram.
P σ (P ) → P → 0
↓ ↓
P σ (M) → M → 0.
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